Unique Paths II - Array - Medium - LeetCode
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Unique Paths II - Array - Medium - LeetCode

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  • 1The robot can only move down or right in an m x n grid to reach the bottom-right corner.
  • 2Obstacles in the grid are represented by 1s, while empty spaces are represented by 0s.
  • 3The solution uses dynamic programming with O(n^2) time and space complexity to calculate unique paths.

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Key Insight
AskGif

"The robot can only move down or right in an m x n grid to reach the bottom-right corner."

Unique Paths II - Array - Medium - LeetCode

A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below).

The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).

Now consider if some obstacles are added to the grids. How many unique paths would there be?

An obstacle and empty space is marked as 1 and 0 respectively in the grid.

Note: m and n will be at most 100.

Example 1:

Input: [ [0,0,0], [0,1,0], [0,0,0] ] Output: 2 Explanation: There is one obstacle in the middle of the 3x3 grid above. There are two ways to reach the bottom-right corner: 1. Right -> Right -> Down -> Down 2. Down -> Down -> Right -> Right

public class Solution {
 public int UniquePathsWithObstacles(int[][] obstacleGrid) {
 int r = obstacleGrid.Length;
 int c = obstacleGrid[0].Length;
 var T = new int[r,c];
 
 bool obstacleFound = false;
 for(int i=0;i<r;i++){
 if(obstacleGrid[i][0]==1){
 obstacleFound = true;
 }
 T[i,0] = obstacleFound ? 0 : 1;
 }
 
 obstacleFound = false;
 for(int i=0;i<c;i++){
 if(obstacleGrid[0][i]==1){
 obstacleFound = true;
 }
 T[0,i]= obstacleFound ? 0 : 1;
 }
 
 for(int i=1;i<r;i++){
 for(int j=1;j<c;j++){
 T[i,j] = obstacleGrid[i][j]==1 ? 0 : T[i-1,j]+T[i,j-1];
 }
 }
 
 return T[r-1,c-1];
 }
}

Time Complexity: O(n^2)

Space Complexity: O(n^2)

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sumitc91

Published on 31 October 2020 · 1 min read · 222 words

Part of AskGif Blog · coding

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